--- title: "Step-down chopper RLE load continuity" subject: "Power Electronics" chapter: 4 key: "semester_7/power_electronics:step-down-chopper-rle-200v" source: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-rle-200v site: BCT Engineering --- # Step-down chopper RLE load continuity Power Electronics · Chapter 4: DC chopper - Asked in: 2079 Baishakh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/step-down-chopper-rle-200v - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-4 ## Question A step down chopper has an input of 200V it is feeding on RLE load having R = 2Ω, L = 10 mH, E = 20 V, the chopping cycle has a time period of 1000 µs and transistor is on for 300 µs in each cycle. (i) Find whether the load current is continuous or not. (ii) Find the average load current. ## Short answer Compare m = E/Vs = 0.1 with critical ratio (e^{αT/Ta} - 1)/(e^{T/Ta} - 1) = 0.279. Since m < 0.279, the load current is continuous. Average load current: V0 = kV_s = 60 V, I0 = (V0 - E)/R = (60 - 20)/2 = 20 A. ## Answer The load current of a step-down chopper feeding an RLE load is continuous when $m = E/V_s$ is less than the critical ratio $\dfrac{e^{\alpha T/T_a}-1}{e^{T/T_a}-1}$ (equivalently, when the computed $I_{min} > 0$). **Given:** $V_s = 200$ V, $R = 2\ \Omega$, $L = 10$ mH, $E = 20$ V, $T = 1000\ \mu$s, $T_{on} = 300\ \mu$s. $$ \begin{aligned} \alpha &= \frac{300}{1000} = 0.3, \qquad T_a = \frac{L}{R} = \frac{0.01}{2} = 5\ \text{ms} \\ \frac{T}{T_a} &= \frac{1}{5} = 0.2, \qquad \frac{\alpha T}{T_a} = 0.06 \end{aligned} $$ #### (i) Continuity of load current $$ \begin{aligned} m &= \frac{E}{V_s} = \frac{20}{200} = 0.1 \\ \frac{e^{0.06}-1}{e^{0.2}-1} &= \frac{1.061837-1}{1.221403-1} = \frac{0.061837}{0.221403} = 0.2793 \end{aligned} $$ Since $m = 0.1 < 0.2793$, **the load current is continuous**. Cross-check with the current limits: $$ \begin{aligned} I_{max} &= \frac{V_s}{R}\cdot\frac{1-e^{-0.06}}{1-e^{-0.2}} - \frac{E}{R} = 100\times\frac{0.058235}{0.181269} - 10 = 22.13\ \text{A} \\ I_{min} &= \frac{V_s}{R}\cdot\frac{e^{0.06}-1}{e^{0.2}-1} - \frac{E}{R} = 100\times0.27929 - 10 = 17.93\ \text{A} \end{aligned} $$ $I_{min} = 17.93$ A $> 0$, so the current never falls to zero. #### (ii) Average load current $$ \begin{aligned} V_0 &= \alpha V_s = 0.3\times200 = 60\ \text{V} \\ I_0 &= \frac{V_0 - E}{R} = \frac{60-20}{2} = 20\ \text{A} \end{aligned} $$ ```text i0 (A) 22.13| /\ /\ /\ | / \ / \ / \ 17.93| / \____/ \____/ \__ |__________________________________ t 0 0.3 1 1.3 2 (ms) ``` **Answer:** (i) continuous ($m = 0.1 < 0.279$); (ii) average load current $I_0 = 20$ A (current swings between 17.93 A and 22.13 A). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The load current is continuous if: A. E/Vs is less than (e^(aT/Ta)-1)/(e^(T/Ta)-1) B. E/Vs is greater than that ratio C. Vs is less than E D. alpha is greater than 0.5 Answer: A. Continuity needs m below the critical ratio, equivalent to Imin > 0. (Key point: Continuity criterion) 2. For Vs = 200 V, E = 20 V, R = 2 ohm, L = 10 mH, T = 1 ms, Ton = 0.3 ms, what is the result? A. m = 0.1 > 0.0793, so the current is discontinuous B. m = 0.1 < 0.2793, so the current is continuous C. m = 0.3 < 0.5, so the current is discontinuous D. m = 0.2793 < 0.1, so the current is continuous Answer: B. Ta = 5 ms, T/Ta = 0.2, aT/Ta = 0.06 and the ratio is 0.2793, bigger than m = 0.1. (Key point: Applying the test) 3. What is the average load current? A. 30 A B. 60 A C. 20 A D. 17.93 A Answer: C. V0 = 0.3 x 200 = 60 V; I0 = (60 - 20)/2 = 20 A. (Key point: Average current) 4. Which pair of current limits cross-checks continuity? A. Imax = 17.93 A, Imin = 22.13 A B. Imax = 20 A, Imin = 0 C. Imax = 100 A, Imin = 10 A D. Imax = 22.13 A, Imin = 17.93 A (positive) Answer: D. Imin > 0 shows current never falls to zero, and the average 20 A lies between the limits. 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